Airgun Thermodynamics

I don't know your answer, but do some googling on this for some neat reading.

In the gun with the pre-charge cylinder the gases are expanding. Both chambers never acquire that equilibrium before the slug starts to move. And the valve closes before you would reach your 50-50 state. This or each consecutive shot would loose a lot of of it velocity.

All true, so let me give a simpler case. These guns are often filled from a scuba tank. Scuba tank at 3000 psi, large volume. The reservoir in the tank is a small volume, at atmospheric pressure. Open the valve, the reservoir and the tank come to equilibrium, at a pressure a bit less than 3000 psi. People have claimed that if this is done too quickly, the reservoir becomes hot, and can detonate if there are oils inside.

A simple experiment, I just don't have the materials to try. It is "common knowledge" in airgunning circles that if you fill the reservoir too fast, it gets hot.

I don't understand the thermodynamics involved.

Really? Have you observed this phenomon yourself?

I think that this is what happens. When the pressurized air from the large cylinder is directed into a smaller vessel, there would be two things happening at the same time. One...the air being compressed will heat up. Two, the air coming into the low pressure vessel from the supply cylinder will drop the temperature of the incoming volume of air dramatically. The volume of cold air exceeds the volume of heating air by many times. If it mixes (and why would it not?) there will be a chilling effect which would more than cancel out the heating effect. I think very quickly, there will be an equilibrium, and the temperature of the air in both the two pressure vessels will become ambient. I think it is likely that the smaller vessel will be very slightly warmer due to the fact that there was a unit of ambient pressure air that got compressed, and you will likely be able to detect that from the outside.

The idea of pressurizing fast enough to ignite any oil in the small pressure cylinder seems unlikely to me. There would have to be a really stratified charge with almost no mixing. Say, with a rubber internal diaphram.

None of my air hammers heat up in operation. In fact, the opposite...they frost up. They are also always soaked in oil, and I have never observed any dieseling. Ever.

Absolutely correct! As pointed out by 34.5 below P1 V1/T1 = P2 V2/T2. In the PCP P2<P1, That is countered by V2>V1 but in all cases there is some loss in the transfer and the net is T2<T1. You can also look at the problem as PV=nRT. V and R are constants in the small chamber. The change in P is only due to a change in n and no "work" is performed.

In the case of the spring loaded system the chamber and piston is one volume that changes radically when the spring moves the piston. The value of n does not change in this case but PV does. The "work" performed by the moving piston show up as an increase in T. see

http://en.wikipedia.org/wiki/Work_(thermodynamics)

Scuba air, mixtures or pure oxygen for breathing may not contain any oil at all. Remember, this goes to the lungs and can suffocate the user. The scuba tanks I have seen filled up are filled with a multi- stage oil free compressor.

When they are "full", the 3000 psi is easily obtained. From there you have the regulator(s) straight to the mouth of the diver, above atmospheric pressure.

I'm talking about contamination of the reservoir or firing chamber with oil, not the air.

Nor did I say it did. It was presumably the smaller tank, the "receiver" which might have some oil in it, and if so, such oil "might detonate". 3000 psi will do that I think! thats a 200 to 1 compression. Diesel engines only have, what, a maximim of 16 to 1 compression, and usually much less.

However in this example, one unit being compressed will heat up, and will mix with a little over 200 units of air being decompressed. And air becomes cold when it decompresses. That mixing of hot and cold will prevent the smaller, presumably oil contaminated cylinder from blowing up.

When the pressure is used to drive a piston, you have a second pressurized mass of air to worry about...the air between the piston and the pellet. Oil on the cylinder walls could easily ignite there, because there is no mixing.

This technique is quite commonly used in a piston fire starter, marketed in most outdoor stores. Works better than you think it would.

The oil in air comment was made by the poster I was replying to

The piston example is pretty simple. Filling from an already compressed source does not appear to be so.

I have never noticed this in my gun (Falcon 19 ms PCP), and when we buy these guns we are always told not to put any oil or grease into the air chamber, as this could possibly catch fire or even blow up! When I fill my guns air chamber with air, the temperature of the air chamber remains the same.

Well, I observed dieseling in piston types. Did not destroy anything, but accuracy was thrown way off. I figure, that at first the pellet is resisting to engage to its grooves, while the piston accelerates. That compresses the air, and raises its temperature. That is classic dieseling.

Pressure chamber types do not have this mechanism.

This has been 'gone into' several times on CR4 - one notable one being a Challenge Question.

Essentially, you apply P1 V1/T1 = P2 V2/T2, to both interconnected volumes.

If there is a compression ignition substance in the mix, and the T2 is high enough, and the concentration is adequate, it will 'detonate'. Then you have a new P1 to match 'backwards' to ascertain if there is a reaction of meaningful proportion via the restriction of the connecting gallery, on said piston and spring.

So not actually complex thermodynamics, more a case of applying the gas laws, and a bit of fluidics, in a functional time/flow sequence.

Sorry, this does not help at all, it's just the ideal gas law. There is nothing in the ideal gas law which deals with a gas getting hotter because of compression.

The piston example is easily calculated, as I originally said, but the ideal gas law does not enter into the equation of how much heat is generated at all.

Also, my question does not involve a spring or piston.

"There is nothing in the ideal gas law which deals with a gas getting hotter because of compression".

Are you sure you meant to type that?

Most definitely. The ideal gas law only describes the relationship of an ideal gas between pressure, temperature and volume at equilibrium.

Let's say you have a cylinder of volume V, sealed by a piston. The piston is forced down to 1/2V. The ideal gas law says that at equilibrium, the pressure will now be double, if the temperature is equal before and after. It does not say anything about how the temperature is changed.

You cannot calculate the rise in temperature for the scenario above using the ideal gas law alone. If the cylinder is insulated, pressing the piston down compressing the gas causes it to heat up, and come to a new temperature T. If you measure this temperature, you can calculate the pressure. If you measure the pressure, you can calculate the temperature.

You can calculate the temperature by knowing how much energy was input into the compression, by knowing the specific heat of the gas etc., but not by the ideal gas law.

I

I now realize I was completely wrong in my above statement. You can use the ideal gas law to calculate the temperature increase. My apologies.

34point5:

Ok, I see how the ideal gas law can be used to calculate the example of a piston compressing a volume of air.

I still don't see how it can be used to calculate temperatures when pressurizing a container from another.

If you would be so kind as to show the calculations for the following simple example:

Two tanks are connected by a valve. Each tank is perfectly insulated, both from each other and the outside world. Each vessel is 24.5 liters in volume. T for both is 298K. Ptank1=20 atm, Ptank2=0 atm.

The valve is opened, then shut when Ptank1=Ptank2. What is the final temperature in tank1 and tank2?

Yar, well, you can do your own math. Besides, Yusef has given you the main clues here.

It doesn't really matter if you are actually talking about refilling a cylinder, or the instant charge introduced behind a ball, or pellet, (unmoving due to friction or inertia). As I mentioned time, so velocity, is a distinguishing component.

Or to all intents and purposes, 'instant' compression of the residual charge, is no different if a piston does it, or more gas does it. This is why paint-ball cylinders and SCUBA tanks are recharged relatively slowly and left to 'temperature equalise' for a period of time. Or when they cool from a 'hot fill' the pressure is significantly less.

The other main difference between this in the real scenario, and your example; is the charging volume (cylinder or piston) is many times greater than the volume being charged.

For the purposes of the 'gas math', you can regard tank 1 as infinite. For the purposes in the recharging real world, tank 1 is infinite until it drops to the regulator setting.

In the case of your compressed air rifle? - from the 'instant' the bobbin unseats, introducing a metered volume into the breach/gallery volume; the 1 atmosphere therein is compressed 'instantly'.

As there is no time for heat to be lost, or move, you can regard this as 'purely adiabatic'. As such, there will be a temperature gradient in the gas. Hot at the pellet to cold at the gallery port.

It is the hot at the pellet that can ignite any fuel present. Once the pellet moves, that hot is again expanding and cooling, whist the balance of the charge, also expanding, flushes any heat momentarily acquired by breach surfaces.

The total effect (over time) is cooling of the device (weapon or pneumatic tool) by the P V /T law from expansion back to standard atmospheric pressure.

I.e. the charge acts much like a 'cold piston' and 'instant' has a great deal to do with 'heat gradients'.

Making sense yet?

It seems that you have a grasp of the problem, as you have stated what I already know.

Thanks, the idea was to present the 'new' in the context of your knowledge as demonstrated.

"New" being; just model the 'instant states' with the applicable gas laws - so come to understand the existence of gradients.

Or, should I interpret your comment as '2nd dismissal'? Making it twice now, you have been reading my posts and dismissing the content by filtering through 'I already know that!' - as opposed to looking for the 'new' clues married into your understanding?

just wondering

I appreciate everyone's comments.

Unfortunately, no one has added anything that I have not thought of. Yusef gave the most complete description of my thoughts, but no quantitative way of calculating the heat rise.

Likewise, your suggestions reflect the way I would approach the calculations and are appreciated, but have not provided me with anything new. Sorry If you took offence.

I even tried to refocus the discussion by showing a very simple model of what I was looking for. Your next email basically rehashed what I had discussed in my original post.

As I said, I am looking for a way of calculating the heat rise when pressurizing one vessel from another. If it is a simple formula, I was hoping that someone would propose it.

My thoughts are that any temperature increase in the charging cylinder are due to an exothermic reaction of contaminates in the system. It's just that whether or not it is a detonation or simple combustion depends upon how the 'fuel' is initially distributed in the tank undergoing the filling as Yusef pointed out. A more apt question would be: how much oil contamination was there given a certain temperature rise?

Otherwise, when simply opening a valve between a high pressure and low pressure vessles, there should be a negligble temperature change assuming both tanks were around the same initial temperature; identical gasses mixed, no work done as Regsoft and 34.5 expanded upon